19 SUPPORT OF A MEASURE. MEASURES WITH COMPACT SUPPORT 213
(ii) If H is any majorized family of positive measures and v = sup H
(13.15.4), then Supp(v) is the closure of the union of the supports of the measures
/ieH.
This follows immediately from (13.16.1) and (13.15.9), applied to the
characteristic function of a relatively compact open set (having regard to
(13.15.4)).
The complement of Supp(/j) can also be defined as the largest of the interiors
of /^-negligible sets. Hence Supp(//) is the intersection of the closures of all
sets on which the measure /x is concentrated (13.18). But it should be realized
that two disjoint measures can have the same support: for example, an atomic
measure on R, concentrated on a denumerable dense set, and such that each
point of this set has nonzero measure (13.18.8), has the same support as
Lebesgue measure.
(13.19.3) For a measure ju on X to be such that every continuous complex-
valued function on X is n-integrable, it is necessary and sufficient that SuppGu)
should be compact. The mapping f\—> /*(/) is then a continuous linear form on the
Fr&chet space #C(X) (12.14.6), and conversely every continuous linear form on
«c(X) is of this type.
If ju has compact support S, then X — S is ju-negligible, hence/<ps is equal to
/almost everywhere. Sincefys is measurable (13.9.6) and \f<ps\ is bounded
above by a multiple of (ps (3.17.10), it follows that/ips, and hence also/, is
integrable.
Let us show conversely that if S = Supp(/z) is not compact, then there
exist continuous real-valued functions /^O such that \/i\*(f) = +00. By
hypothesis (3.18.3) there exists an increasing sequence of relatively compact
open sets Un in X such that Un c Un+1 and the Un cover X. Since X is not
compact, we have Un ^ X for all n. We now define inductively a sequence (an)
of points of X and a sequence (Vn) of relatively compact open sets, as follows:
#! e S; Y! is a neighborhood ofai;an+l belongs to the intersection of S and
the complement of the union of On and the Vk with k ^ n (by hypothesis, this
intersection is not empty); VB+1 is a relatively compact neighborhood of
an+l such that Vn+1 does not intersect 0,, nor any V* with k ^ n, For each n9
let gn be a continuous mapping of X into [0, 1] which takes the value 1
at an and the value 0 throughout X — Vn (4.5.2). The hypothesis implies
that |Ju|(^n)>0. Put fn=9nl\V\(9^ so that |^|(/„) = 1. Then the function
00
/= y fn is everywhere finite and continuous (because every point
M=l
HUNT LIBflftBY
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IBHage of /x under TT is Lebesgue measure A on I.